--%>

Schrodinger equation with particle in a box problem.

Three dimensional applications of the Schrodinger equation are introduced by the particle-in-a-box problem.

So far only a one-dimensional problem has been solved by application of the Schrodinger equation. Now the allowed energies and the probability functions for a particle that is free to move in three dimensions are deduced. A molecule of a gas enclosed in a cubic container provides a specific example that is dealt with in the section after the general procedure has been developed.

For any three-dimensional problem, the potential energy is, general, a function of three coordinates. For a cubic potential box, the Cartesian coordinates are convenient. The differential equation that must be solved is now the Schrodinger equation in three dimensions.

1310_Particle in a box.png 

For a "cubic box," the potential function can be expressed in terms of separate x, y, and z components,
98_Particle in a box1.png 

Each of the potential function components for a "particle-in-a-box" is like the one-dimensional potential for a "particle-on-a-line".

For three-dimensional systems, the solution function ψ depends on the three coordinates necessary to locate a point in space. It is often profitable to try to separate such systems into parts, with each part involving only one coordinate. On the basis we try the substitution

ψ (x, y, z) = Ø(x)Ø(y)Ø(z)

Substitution of (2) from (1) gives

1350_Particle in a box2.png 

Division by Ø(x)Ø(y)Ø(z) gives

1180_Particle in a box3.png 

For the equation to be satisfied for all values of x, y and z, each term on the left must equal a component of ε, and we can write

ε = εx + εy + εz

The Schrodinger equation can then be broken down into three identical equations of the type

1394_Particle in a box4.png 

Or

578_Particle in a box6.png 

These equations are identical to that written for one-dimensional problem. The solution to the three-dimensional cubic-box problem is therefore

ψ =  Ø(x)Ø(y)Ø(z)

With

1809_Particle in a box7.png

   Related Questions in Chemistry

  • Q : Value of molar solution Select the

    Select the right answer of the question. Molar solution contains: (a)1000g of solute (b)1000g of solvent (c)1 litre of solvent (d)1 litre of solution

  • Q : What is cannizaro reaction? Explain

    Aldehydes which do not have  -hydrogen atom, such as formaldehyte and benzaldehyte, when heated with concentrated (50%)alkali solutio

  • Q : Explain the molecular mass with respect

    During the formation of polymers, different macromolecules have different degree of polymerisation i.e. they have varied chain lengths. Thus, the molecular masses of the individual macromolecules in a particular sample of the polymer are different. Hence, an average value of the molecular mass is

  • Q : What are homogenous catalyst? Give few

    When a catalyst mixes homogeneously with the reactants and forms a single phase, the catalyst is said to be homogeneous and this type of catalysis is called homogeneous catalysis. Some more examples of homogeneous catalysis are:    SO2

  • Q : Reaction of calcium carbonate Give me

    Give me answer of this question. What is the volume of 0.1NHcl required to react completely with 1.0g of pure calcium carbonate : (Ca= 40, C= 12 and o = 16 ) (a)150cm3 (b)250cm3 (c)200cm3 (d)100cm3

    Q : Explain gels and its various categories.

    Certain sols have the property of setting to a semi-solid, jelly-like form by enclosing the entire amount of liquid within itself when they are present at high concentrations. This process is called gelation and colloidal systems with jelly-like appearance are known as gels. Some common examples

  • Q : Neutralization of sodium hydroxide How

    How much of NaOH is needed to neutralise 1500 cm3 of 0.1N HCl (given = At. wt. of Na =23): (i) 4 g  (ii) 6 g (iii) 40 g  (iv) 60 g

  • Q : Theory of three dimensional motion

    Partition function; that the translational energy of 1 mol of molecules is 3/2 RT will come as no surprise. But the calculation of this result further illustrates the use of quantized states and the partition function to obtain macroscopic properties. The partition fu

  • Q : BASIC CHARACTER OF AMINES IN GAS PHASE,

    IN GAS PHASE, BASICITIES OF THE AMINES IS JUST OPPOSITE TO BASICITY OF AMINES IN AQEUOUS PHASE .. EXPLAIN

  • Q : Molar mass Select the right answer of

    Select the right answer of the question. Which is heaviest: (a)25 gm of mercury (b)2 moles of water (c)2 moles of carbon dioxide (d)4 gm atoms of oxygen